Poles and Bode Plots

I have three questions that have been troubling me for a long while:

1. We say that, in a Bode plot, there is a drop in gain of 20 dB per decade whenever a pole is encountered. But aren't poles defined as the values of $s$ which make the transfer function infinity? So why doesn't the gain go up at this point instead of going down?

2. Physically what happens when we feed a system with a pole frequency?

3. Also, consider a transfer function $1/(s+2)$. The system has pole at $s=(-2+j0)$. That is, for the pole, $\sigma=-2$ and $\omega=0$. But when we apply a sinusoidal signal to its input and draw the Bode plot, why do we say that there is a pole at 2 rad/sec (even though, for the pole, $\omega=0$ and $\sigma =-2$)?

• Do you know the meaning of the "pole frequency"? It is a frequency identical to the length of the vector from the origin to the pole location (Pythagoras rule). In case of a real pole the pole frequency is identical to the negative real part (-sigma). Hence, it is not possible to excite any circuit with its pole frequency. It is just an artificial - but very helpful tool . – LvW Mar 12 '15 at 15:16
• @LvW: That frequency is usually called the natural frequency. The pole frequency is determined by the imaginary part of the pole. – Matt L. Mar 12 '15 at 16:02
• Matt L., sorry but I disagree. Ill look for some references. – LvW Mar 12 '15 at 16:04
• Matt L., I am afraid, there is a difference in the terminology between Germany and the US. I think, I must agree that in your country the parameter we call "pole frequency" is known as "natural frequency". Sorry. – LvW Mar 12 '15 at 16:12
• @Matt L., I am happy to tell you that I am not completely "out of the track": There is a book on filter techniques "Analog and Dig. Filters" (Harry Y.F.Lam, Bell Inc.) in which the magnitude of the pole location (distance from the origin) also is called "pole frequency". Good to know, but we always should be cautiuous while using such keywords. – LvW Mar 13 '15 at 14:09

Bode plot is not a graph that plots the transfer function ($H(s)$) against $s$. $H(s)$ is a complex function and its magnitude plot actually represents a surface in Cartesian coordinate system. And this surface will have peaks going to infinity at each poles as shown in figure:

Bode plot is obtained by first substituting $s= j\omega$ in $H(s)$ and then representing it in polar form $H(j\omega) = |H(\omega)|\angle\phi(\omega)$. $H(\omega)$ gives the magnitude bode plot and $\phi(\omega)$ gives the phase bode plot.

Bode magnitude plot is the asymptotic approximation of the magnitude of transfer function ($|H(\omega)|$) vs logarithm of frequency in radians/sec ($\log_{10}|\omega|$) with $|H(s)|$ (expressed in dB) on y-axis and $\log_{10}|\omega|$ on x-axis.

Coming to the questions:

1. At poles, the complex surface of $|H(s)|$ peaks to infinity not $|H(\omega)|$.

2. When a system is fed with pole frequency, the cosponsoring output will be having the same frequency but amplitude and phase will be changing. The value can be determined by substituting the frequency in radians/sec in $|H(\omega)|$ and $\phi(\omega)$ respectively.

3. A pole at -2 rad/sec and 2 rad/sec have the same effect on $|H(\omega)|$. And our interest is in the frequency response. So we need only positive part of it.

• Nice answer and I love that you took the time to format it nicely! +1 – Null Mar 12 '15 at 15:04
• I can't follow. First, H(s) itself does not represent a surface as you show; instead, it has a complex value at each (complex) s. What you display is probably the absolute value (magnitude) |H(s)|, or maybe the real part, real(H(s)). As for what you say in the first paragraph below the image: If real(H(s)) and/or imag(H(s)) go to infinity, then the magnitude, |H(s)|, also goes to infinity. How could it not? – Christopher Creutzig Mar 12 '15 at 15:36
• @ChristopherCreutzig The graph shown is a 3D plot. real part of 's' on x-axis, imaginary part of 's' on y-axis and magnitude of H(s) on z-axis. but I can see that there are some confusions. Let me make an edit. – nidhin Mar 12 '15 at 15:50
• I got that part. My complaint is that the graph is not of H(s), since it is simply impossible to plot a complex function of a complex parameter this way (when using fewer than four dimensions). The surface shown is that of |H(s)| and should not be called a surface (plot) of H`. – Christopher Creutzig Mar 12 '15 at 17:44
• @Christopher now I got you. I was using the words in a quite confusing way. Hope that I made it clear this time. – nidhin Mar 12 '15 at 18:50

When trying to understand transfer functions, I think the "rubber-sheet analogy" is very useful. Imagine an elastic rubber-sheet covering the complex $s$-plane, and imagine that at every zero of the transfer function the sheet is tacked to the ground, and at every pole there is a literal thin pole pushing the rubber-sheet up. The magnitude of the frequency response is the height of the rubber-sheet along the $j\omega$-axis.

1. From the above analogy, of course the gain goes up towards the pole. But moving away from the pole, the contribution of the pole makes the transfer function go down (e.g. going towards the next zero). Imagine the simple system you gave as an example in your third question. It has a real-valued pole at $s_{\infty}=-2$, and - due to this pole - it also has a zero at $s_0=\infty$. So moving away from the pole with increasing frequency, the transfer function goes down because the rubber-sheet is tacked to the ground at infinity. Mathematically, this is also easy to see: $$H(s)=\frac{1}{s+2}\Longrightarrow \left|H(j\omega)\right|^2=\frac{1}{\omega^2+4}=\frac14\frac{1}{\left(\frac{\omega}{2}\right)^2+1}$$ In decibels we get $$10\log_{10}\left|H(j\omega)\right|^2=-10\log_{10}(4)-10\log_{10}\left[\left(\frac{\omega}{2}\right)^2+1\right]\tag{1}$$ For $\omega\gg 2$ the second term on the right-hand side of (1) can be approximated by $$-10\log_{10}\left(\frac{\omega}{2}\right)^2=-20\log_{10}(\omega/2)$$ which is a straight line with a slope of $-20\,\text{dB}$ per decade.

2. When you excite a system with a signal corresponding to one of its poles, then this input signal is "amplified" compared to input signals with other frequencies. Note, however, that for a stable system the output signal will always decay. E.g. if you excite the system with transfer function $H(s)=\frac{1}{s+2}$ with an input signal $x(t)=e^{-2t}$, then the output will be $y(t)=te^{-2t}$, where the factor $t$ corresponds to the system's "amplification" of the input signal. However, the exponential factor will make the signal approach $0$ for large values of $t$.

3. In short, we don't say that there's a pole at $2$ rad/s, because there isn't. What is indeed the case is that the cut-off frequency is determined by the real part of the pole, i.e. the starting point of the line with negative slope in the Bode plot is determined by the value $2$. This is the example I gave in point 1 above, where the straight line approximation with $-20$ dB per decade is valid for $\omega\gg 2$. The value $2$ is not determined by the pole frequency (which is zero) but by the real part of the pole.

• I've heard that analogy before and I think it's the best one for understanding the concept. And thanks for taking the time to format your answer nicely! +1 – Null Mar 12 '15 at 15:05

The graph shows the difference between the natural frequency in the complex $s$-plane (infinite) and the corresponding magnitude peak along the $j\omega$ axis which can be observed during measurements: The graph belongs to a natural frequency of $\omega_p=1000$ rad/s and a pole quality factor $Q_p=1.3$ (which is a measure of the observable gain peaking). This plot visualizes a 2nd-order Chebyshev characteristics with 3 dB ripple in the passband.

The "s" in your equations is the constant in the function exp(s*t). So, when s is a real number, this time function is an exponentially growing or falling function. Your example with s=-2 is an exponentially falling function. For any pole "number", the output will grow when you apply an input at that "number". If you apply an exponentially falling signal to your example circuit, the output signal will go to infinity. (Note, however, that it is not possible to generate a signal that is always exponentially falling, because such a signal is very large at times in the past). When you talk of frequencies like 2 radians/sec, you are speaking of poles at j*2, not 2, so those signals are sinusoidal. It is possible to generate signals that are sine waves (at least for a pretty long time). You will get infinities out if you apply this sine wave signal to a system with a pole at +-j*2, but not if you apply it to a system with a pole at -2.

• Since u have not answered his question this should be a comment – Pedro Quadros Mar 12 '15 at 14:18